Can prime quadruplets be adjacent Can prime decade's ( 101, 103, 107, 109 ) be adjacent? And if not, why not?
I.e.:
{101, 103, 107, 109} --- {111, 113, 117, 119}
(I'm aware the second set isn't actually a valid prime decade, just using as an example of what I mean)
For any explanation of why (or not), please keep it very simple!
Thanks.
 A: No. Your numbers are of the form $10n+1$, $10n+3$, $10n+7$, $10n+9$, $10n+11$, $10n+13$, $10n+17$ and $10n+19$ for some $n\geq0$.
One of $10n+4$, $10n+5$ and $10n+6$ must be divisible by $3$. If it's $10n+4$, then $10n+7$ is divisible by $3$, so not prime. If it's $10n+6$, then $10n+9$ is divisible by $3$ so is not prime. Finally, if it's $10n+5$, then $10n+11$ is divisible by $3$, so is not prime.
Another way of reading this argument; if $10n+1$, $10n+3$, $10n+7$ and $10n+9$ is a prime decade, then $10n+5$ must be divisible by $3$, by the above argument. Two consecutive numbers of the form $10n+5$ are never both divisible by $3$, because their difference is $10$, so there cannot be two consecutive prime decades.
A: The first number of such a decade is the first number of a prime twin pair. These are always of the form $6k\pm1$, so the first number of such a decade must be of the form $6k-1$. So diffferent decades must be a multiple of $6$ apart.
A: If $p > 3$ is prime, then $p$ must be of the form $3n+1$ or $3n+2$. In the first case, $p+2=3n+3$ is divisible by $3$; in the second case, $p+4=3n+6$ is divisible by $3$.
So $p,p+2,$ and $p+4$ can't all be prime.
